Torque, moments, and angular momentum
Cross product and torque
In all of the torque problems I've done so far in the physics playlist, we only just figured out the magnitude of torque, frankly because that's what normally matters, but torque is actually a vector and it's direction can be found. And that is because torque is defined as the cross product between the radial distance from your axis of rotation and the rotational force being applied. So these are both vectors. So let's take a look at how I taught you vectors the first time, and then I'll show you how that's really the same thing as what we're doing here with the cross product. Except now with the cross product, besides just the magnitude for torque, we're also getting the direction. But then we'll also see that direction is a little bit-- it's just the definition of the direction of torque. I don't know how intuitive it really is. But what did I teach you before about torque? Well, let's say I had some arm, and let's say this could be the hand of a clock or it's pinned down to the wall there. So it would rotate around this object. Let's say it's some distance, r, from the pivot. Let's say that distance is 10. This is the same thing as r, and the magnitude of r is equal to 10. At some distance 10 from the pivot, I apply some force F, and F I will do in yellow. I apply some force F. Let me draw it straight. I apply some force F at some angle. That's my force F. It's also a vector. It has magnitude and direction. Let's say that this is 10 meters, and let's say that I apply a force of 7 newtons. Let me make it more interesting. Let's say I apply a force of square root of 3 newtons. And I just threw that out there because I think the numbers will all work out. And let's say that the angle between my force and the lever arm, or the arm that's rotating-- let's stick to radians this time. Let's say it's pi over 3, but if you need to visualize that, that's 60 degrees. pi over 3 radians is equal to theta. And so just based on what we already know about moments or torque, what is the torque around this pivot? Or how much torque is being applied by this force? And when we learn torque or we learn moments, we realize really the only hard part about these problems is that you don't just multiply the entire rotational force times the distance from the axis of rotation. You have to multiply the component of that force that is actually doing the rotation, or the component of the force that is perpendicular to this rotating arm, or perpendicular to this moment arm. So how do we figure that out? Well, the component of this force that is perpendicular to this arm-- I can visually draw it here. Let's see, it would look something like this. I could draw it there. I could also draw it here, right? This would be the component, or this would be the component that is perpendicular to this rotating arm, and the component that is parallel would be this, but we don't care about that. That's not contributing to the rotation. The only thing that is contributing to the rotation is this component of the force. And what is the magnitude of this vector right here? The component of vector F that is perpendicular to this arm. Well, if this angle-- let me draw a little triangle down here. If this is square root of 3, this is pi over 3 radians, or 60 degrees, and this is a right angle, it's pi over 3. I know it's hard to read. What is this length right here? Well, it's a 30-60-90 triangle, and we know that this length here-- I mean, there's a couple of ways you can think about it. Now that we know trigonometry, we know that this is just the square root of 3 times the sine of pi over 3, or the sine of 60 degrees, and so that equals the square root of 3. Sine of pi over 3, or sine of 60 degrees, is square root of 3 over 2. So the square root of 3 times the square root of 3 is just 3, so that equals 3/2. So the magnitude of this force vector that is perpendicular, the component that is perpendicular to the arm, is 3/2 newtons, and now we can figure out the magnitude of the torque. It's 3/2 newtons times 10 meters. So we know the magnitude of the torque, and I'm being a little bit more careful with my notation right now to remind you that torque actually is a vector, or you can almost view it as, they use this term pseudovector, because it's kind of a-- well, anyway, I won't go into that. So what is the magnitude of the torque vector? Well, it's 3/2 newtons times the distance, and remember, I just drew this vector here just to show you the component. I could just shift the vector here because this is actually where the force is being applied. You could draw that same vector here because you can shift vectors around, so this is also 3/2 newtons and maybe that makes it a little bit clearer. So it's 3/2 newtons times the distance that you are from your pivot arm, so times 10 meters, and so that is equal to what? 15 newton meters. So the magnitude of the torque is 15 newton meters. But all we did now-- and hopefully this looks a little bit familiar. This is what we learned when we learned moments and torque, but all we did now is we figured out the magnitude of the torque. But what if we wanted to know the direction? And that's where the cross product comes in. So what was the definition of the cross product? Cross product: r cross F, that is equal to magnitude of r times the magnitude of F times sine of the smallest angle between them times some vector that is perpendicular to both. And this is really where it's going to help, because all of these right here, these are all scalar quantities, right? So these don't specify the direction. The direction is completely specified by this unit vector, and a unit vector is just a vector of magnitude 1 that's pointing in some direction. Well, look, this cross product, this part of it, the part that just gives us magnitudes, we just calculated that using what we knew before of torques. The magnitude of our force vector times sine of theta, that gave us the component of the force vector that is perpendicular to the arm. And we just multiply that times the magnitude of r, and we got the magnitude of the torque vector, which was 15. We can leave out the newton meters for now. 15, and then its direction is this vector that we specified by n. We can call it the normal vector. And what do we know about this vector? It's perpendicular to both r-- this is r, right-- and it's perpendicular to F. And the only way that I can visualize in our three-dimensional universe, a vector that's perpendicular to both this and this is if it pops in or out of this page, right? Because both of these vectors are in the plane that are defined by our video. So if I'm a vector that is perpendicular to your screen, whatever you're watching this on, then it's going to be perpendicular to both of these vectors. And how do we figure out if that vector pops out or pops into the page? We use the right hand rule, right? In the right hand rule, we take-- r is our index finger, F is our middle finger, and whichever direction our thumb points in tells us whether or not we are-- the direction of the cross product. So let's draw it. Let me see if I can do a good job right here. So if that is my index finger, and you could imagine your hand sitting on top of this screen. So that's my index finger representing r, and this is my right hand. Remember, it only works with your right hand. If you do your left hand, it's going to be the opposite. And then my middle finger is going to go in the direction of F, and then the rest of my fingers are-- and I encourage you to draw this. So if I were to draw it-- let me draw my nails just so you know what this is. So this is the nail on my index finger. This is the nail on my middle finger. And so in this situation, where is my thumb going to be? My thumb is going to be popping out. I wish I could-- that's the nail of my thumb. Hopefully, that makes some sense, right? That's the palm of my hand. That's the other side of my-- and I could keep drawing, but hopefully, that makes some sense. This is my index finger. This is the middle finger. My thumb is pointing out of the page, so that tells us that the torque is actually pointing out of the page. So the direction of this unit vector n is going to be out of the page, and we could signify that by a circle with a dot. And I'm almost at my time limit, and so there you have it: the cross product as it is applied to torque. See you in the next video.